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If $x + \frac{1}{x} = 12$ find the value of $x - \frac{1}{x}$.
Given:
$x + \frac{1}{x} = 12$
To do:
We have to find the value of $x - \frac{1}{x}$.
Solution:
The given expression is $x + \frac{1}{x} = $. Here, we have to find the value of $x^4 + \frac{1}{x^4}$. So, by squaring the given expression and using the identities $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$, we can find the required value.
$(a+b)^2=a^2+2ab+b^2$.............(I)
$(a-b)^2=a^2-2ab+b^2$.............(II)
Let us consider,
$x + \frac{1}{x} = 12$
Squaring on both sides, we get,
$(x + \frac{1}{x})^2 = (12)^2$
$x^2+2\times x \times \frac{1}{x}+\frac{1}{x^2}=144$ [Using (I)]
$x^2+2+\frac{1}{x^2}=144$
$x^2+\frac{1}{x^2}=144-2$ (Transposing $2$ to RHS)
$x^2+\frac{1}{x^2}=142$
Now,
$x^2+\frac{1}{x^2}=142$
Subtracting 2 from both sides, we get,
$x^2+\frac{1}{x^2}-2=142-2$
$x^2-2\times x \times \frac{1}{x}+\frac{1}{x^2}=140$ [Since $2\times x \times \frac{1}{x}=2$]
$(x-\frac{1}{x})^2=140$ [Using (II)]
Taking square root on both sides, we get,
$x-\frac{1}{x}=\sqrt{140}$
Hence, the value of $x-\frac{1}{x}$ is $\sqrt{140}$.
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